[Notation. $a_n \lesssim b_n$ means: there exists some positive constant $C$ s.t. $a_n \le C b_n$.]
A rough'n'ready argument would be:
let
$$c_n=\frac{1}{2 \pi} \int_{-\pi}^{\pi}f(y)e^{-i n y}\, dy$$
and write
$$f(x)=\sum_{n\in \mathbb{Z}} c_n e^{i n x}\quad (1)$$
The decay condition on $c_n$ implies uniform convergence of this series:
$$\lvert c_n e^{i n x} \rvert \le \lvert n\rvert^{-k}\lvert n^k c_n \rvert \lesssim \lvert n\rvert^{-k}$$
and $\sum_{n \in \mathbb{Z}} \lvert n\rvert^{-k}$ is a convergent numerical series. Now differentiate (1) termwise: you get
$$\sum_{n \in \mathbb{Z}}i n c_n e^{i n x}$$
which is again a uniformly convergent series:
$$\lvert i n c_n \rvert \lesssim \lvert n\rvert^{1-k}.$$
So (1) is a uniformly convergent series whose term-by-term derivative is uniformly convergent. This implies that $f$ is differentiable and
$$f'(x)=\sum_{n \in \mathbb{Z}}i n c_n e^{i n x}$$
so that, again by uniform convergence, $f'$ is also continuous.
It really boils down to integration by parts.. the $n$th Fourier coefficient of ${d^k f \over dt^k}$ is given by $(in)^k \hat{f}(n)$, which you prove by integrating by parts $k$ times. If $f$ is $C^k$, then the Fourier coefficients of the continuous function ${d^k f \over dt^k}$ go to zero by the Riemann-Lebesgue lemma, so you get that $|n|^k|\hat{f}(n)|$ goes to zero or that $|\hat{f}(n)|$ is $o({1 \over |n|^k})$.
If you have a small discontinuity in the $k$th derivative, you might get a small error term like $\epsilon {1 \over |n|^{k-{1 \over 2}}}$ just to give an example, and as the discontinuity goes to zero the $\epsilon$ will go to zero... but for any $\epsilon > 0$ it will still be the dominant term if $|n|$ is large enough. So slowly changing the function will gradually make this term go away in one sense, but it will still dominate in another sense regardless of what $\epsilon$ is.
Best Answer
In stated generality this is false. Take some periodic function with behaves like $f(x)=x^{1/3}$ near $0$. Since $f'(x)\notin L^2$, the Fourier coefficients of $f'$ are not in $\ell^2$. In particular, they cannot decay like $1/n$.
The following (adapted from here) is true.
Claim. Suppose $f $ is piecewise differentiable, $f'$ has bounded variation on each "piece" $(a_k,a_{k+1})$ and has nonzero jump discontinuity at some $a_k$. Then the sequence $n^2\widehat f(n)$ is bounded but does not tend to $0$.
Proof: $n^2 \widehat f(n)$ has the same size as $\widehat \mu(n)$, where $\mu$ is the signed measure such that $f'=\int \,d\mu$. (That is, $\mu$ is the weak/distributional derivative of $f'$). By the assumptions on $f'$, the measure $\mu$ consists of absolutely continuous component and some point masses. The Fourier series for the absolutely continuous part has coefficients that tend to $0$, by the Riemann-Lebesgue lemma. The Fourier series for the sum of point masses is a trigonometric polynomial; therefore it is bounded but does not tend to $0$. $\quad \Box$
With some extra machinery, Claim can be generalized:
Claim 2. Suppose $f $ is absolutely continuous, $f'$ has bounded variation and is not continuous. Then the sequence $n^2\widehat f(n)$ is bounded but does not tend to $0$.
Proof goes as above, but the sticky point is to show that $\widehat \mu(n)\not\to 0$. A measure such that $\widehat \mu(n) \to 0$ is called a Rajchman measure. From the survey Seventy years of Rajchman measures by Lyons we find that a Rajchman measure has no atoms (proved by Neder in 1920). But $f'$ is not continuous, which means the measure $\mu$ has atoms. $\quad \Box$